\magnification=\magstep1
\centerline{Mathematics 174 -- Applied Mathematics 2}
\centerline{Final Problem Set}
\centerline{April 27, 2000}
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\noindent
{\it Ground Rules}
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This is an open book, open notes, open library, open Maple,
open SW 335, unlimited time
exam.  If you find something of use in a library book, please footnote
it (this is not necessary for topics covered in our course textbook, 
Haberman).  You must work independently, and you may not consult other
faculty members in working on these problems.  {\it When you have completed
the problem set, please write a brief statement saying that you have abided
by these ground rules, sign it, and turn it in with your solutions}.
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\noindent
{\bf Due:}  No later than 5:00 pm, Friday, May 5.
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\noindent
{\it Directions}
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Do problem I below, and any {\it two} of problems II, III, IV.  
(If you submit solutions for all four problems (or parts of them) 
I will give Extra Credit points.)

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\noindent
I. (40) In this problem, you will see another example of
how Fourier Transforms can be applied to solve PDE.
Consider the following problem for the
{\it linearized Korteweg-deVries equation}
$$\cases{u_t = u_{xxx}&\cr
         u(x,0) = f(x)& \cr}\leqno(1)$$
(This is a simplified version of an equation used to model
the motion of water waves in long, shallow channels like canals -- 
$u(x,t)$ represents the vertical displacement at position $x$
along the canal, at time $t$.  For a long canal, and
a wave that starts due to some disturbance of the
water surface in the middle, there are essentially
no boundary conditions, so we take $-\infty < x < \infty$.)
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\item{A)}  Take the Fourier Transform (with 
respect to $x$) of both sides of the PDE  to find an ODE 
(with independent variable = $t$) satisfied by $U(\omega,t) = {\cal F}[u(x,t)]$
and solve it to find $U(\omega,t)$.  Don't forget to include a 
``constant of integration'' here, which will depend on $\omega$.
Take the inverse transform to give one way find a solution of (1).
\item{B)}  Use the convolution theorem for the Fourier
Transform to show that the solution
$u(x,t)$ can be written as
$$u(x,t) = f(x) * G(x,t) = \int_{-\infty}^\infty f(s) G(x - s,t)\ ds$$
where the function $G$ is given by 
$$G(x,t) = {1\over 2\pi}\int_{-\infty}^\infty \cos(\omega^3t - \omega x)\ d\omega
= {1\over\pi}\int_0^\infty \cos(\omega^3t - \omega x)\ d\omega.$$  
This is called the {\it fundamental solution} -- it is the solution 
corresponding to $u(x,0) = \delta(x)$ (Dirac delta).
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\noindent
By a happy chance, there is a built-in function called {\tt AiryAi} 
in Maple (with an efficient method for computing values) 
that is very closely related to this function:
$$G(x,t) = {{\rm AiryAi}\left({-x\over {\root 3\of {3t}}}\right)\over 
{\root 3\of {3t}}}$$
This is good because the integral representation above is {\it very 
inefficient\/} for evaluating $G(x,t)$ -- it's similar to a very 
slowly-converging alternating series.
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\item{C)}  Use Maple to generate plots of the solution of (1) if 
$$f(x) = \cases{\sin(\pi x) & if $x \in [0,1]$\cr
                0 & otherwise\cr}$$
and describe what the solution $u(x,t)$ appears to be doing as $t$ increases.
Note: there are several different ways to think about computing the 
convolution formula from part B (either integrating directly, or 
by using the FFT in a suitable way to approximate).  Explain how you are 
doing that step.
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\noindent
II. (30) Consider a Sturm-Liouville form differential equation:
$${d\over dx}\left(p(x){dy\over dx}\right) + q(x) y = 0,
\leqno(2)$$
and assume $p(x) > 0$ for all $x \in [a,b]$ (as in a regular Sturm-Liouville
eigenvalue problem).  
In this problem, we will study the {\it zeroes} of solutions of (2), and
prove a result called the {\it Sturm Separation Theorem} -- part B below.
Let $y_1,y_2$ be any two solutions of (2). The {\it Wronskian} 
of $y_1, y_2$ is defined to be:
$$W[y_1,y_2](x) = \det\pmatrix{y_1(x)&y_2(x)\cr
                               {dy_1\over dx}(x)&{dy_2\over dx}(x)\cr}$$
\item{A)} Show that $W[y_1,y_2](x) = C/p(x)$ for some constant $C$. ({\bf Hint:} What
is the derivative ${d\over dx}W[y_1,y_2](x)$?)  Use this to 
show that if the Wronskian of $y_1$ and $y_2$ is zero at any 
$c \in [a,b]$, then $y_1$ and $y_2$ are {\it linearly dependent}.  You
will need to use the fact that for all $c \in [a,b]$, and all $A,B\in {\bf R}$, 
there is a {\it unique}
solution of (1), together with the initial conditions $y(c) = A$, 
${dy\over dx}(c) = B$.
\item{B)}  Let $y_1,y_2$ be two linearly independent solutions of (2).  
Show that between any two consecutive zeroes of $y_1$ in $[a,b]$, there
is precisely one zero of $y_2$ as follows:  Assume $x_1 < x_2$ are the
consecutive zeroes of $y_1$, but that $y_2$ is not zero on the 
interval $[x_1,x_2]$.  Consider the function ${y_1\over y_2}$ and 
apply the ``most valuable theorem'' (MVT) of calculus.
\item{C)}  What does this result say about the zeroes of the 
Bessel functions $J_n(z)$ and $Y_n(z)$?  

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\noindent
III.  (30) In this problem, you will see a proof of the {\it Uncertainty Principle},
which implies (among other things) that signals {\it cannot} be simultaneously
arbitrarily sharply time-limited and frequency-limited. 
 The famous Heisenberg uncertainty 
principle from quantum mechanics is also derived from this fact about
Fourier analysis.  Throughout this problem, we will use the 
following notation (similar to our inner-products from Sturm-Liouville
theory):  Let $f(x), g(x)$ be ``nice'' signals (i.e. piecewise
smooth, and in $L^1({\bf R})$).  We write
$$\langle f, g \rangle = \int_{-\infty}^\infty f(x) \overline{g(x)}\ dx
\leqno(3)$$
where $\overline{g(x)}$ is the complex conjugate of $g(x)$  (i.e. if
$g(x) = u(x) + iv(x)$, then $\overline{g(x)} = u(x) - iv(x)$).
We also write $\|f\| = \sqrt{\langle f,f\rangle} = 
\sqrt{\int_{-\infty}^\infty |f(x)|^2\ dx}$ (called the 
$L^2$-norm of $f$).
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\item{A)}  First, we need an important fact called the {\it Plancherel
identity}.    Show using the Fourier Inversion Theorem that
$$\langle {\cal F}[f], {\cal F}[g]\rangle = {1\over 2\pi}\langle f,g\rangle,$$
where the left-hand side is the inner product from (3) above, but for
functions of $\omega$.  From
this deduce the Plancherel identity:
$${1\over 2\pi} \|f\|^2 = \|{\cal F}[f]\|^2.$$
\item{B)}  We assume all signals are ``centered'' at $x = 0$.
The statement of the Uncertainty Principle will involve
a measure of the ``effective width'' of a signal and its Fourier 
transform.  For the effective widths we will use the following
definitions.  The effective width of $f$ is 
$$\eqalign{
W_f &= {\| x f(x) \|\over \|f(x)\|}\cr
&=\left[{1\over \|f\|^2}\int_{-\infty}^\infty x^2 |f(x)|^2\ dx\right]^{1\over 2}\cr}$$
(this is related to a {\it variance} as defined in statistics; if that
doesn't mean anything to you, ignore it!!)
Similarly, we define the effective width of the Fourier transform to be
$$\eqalign{
W_F &= {\| \omega {\cal F}[f](\omega)\|\over \|{\cal F}[f](\omega)\|}\cr
&=\left[{2\pi\over \|f\|^2} 
\int_{-\infty}^\infty \omega^2 |{\cal F}[f](\omega)|^2\ d\omega\right]^{1\over 2}\cr}$$
Compute the effective width $W_f$ for $f(x) = e^{-a x^2}$ (scaled Gaussian),
and $W_F$ for the Fourier transform of the Gaussian.  What is the
product $W_f W_F$ in this case?
\item{C)}  The Uncertainty Principle is the statement:
$$W_f W_F \ge {1\over 2}$$
for all {\it real} signals $f$.  (Hence, in particular, $W_f$ and $W_F$ cannot
both be small!) Prove this statement by 1) recognizing that
$W_F$ is equal to ${\|f'\| \over \|f\|}$, 2) applying
the Cauchy-Schwarz inequality: 
$$\|x f(x)\| \| f'(x)\| \ge \left|\int_{-\infty}^\infty x f(x) f'(x)\ dx\right|,
$$
then 3) working with the integral on the right above.  (You may assume
$x (f(x))^2 \to 0$ as $x\to \pm\infty$.)

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\noindent
IV. (30) In this problem, you will work through most of the
steps of a proof of the {\it Shannon-Nyquist
Sampling Theorem}, which is the following statement:
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\proclaim Sampling Theorem. Let $f$ be piecewise smooth and in $L^1({\bf R})$.
Assume that ${\cal F}(\omega) = 0$ for all $|\omega| \ge L$.  Then
$$f(x) = \sum_{n=-\infty}^\infty f\left({n\pi\over L}\right)
{\sin(Lx - n\pi)\over Lx - n\pi}.\leqno(4)$$

\noindent
The meaning of this is that if $f$ is {\it frequency limited}
(that is, ${\cal F}[f](\omega) = 0$ for all $|\omega| \ge L$), 
then we can reconstruct $f$ {\it exactly} (in principle) from sampled values 
$f\left({n\pi\over L}\right)$, provided that the sampling interval
is ${\pi\over L} = {2\pi \over 2L}$.  Any sampling interval
$\Delta x < {2\pi \over 2L}$ would also work.

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\item{A)}  By the Fourier Inversion Theorem,
$$f(x) = \int_{-\infty}^\infty {\cal F}[f](\omega) e^{-i \omega x}\ d\omega
=  \int_{-L}^L {\cal F}[f](\omega) e^{-i \omega x}\ d\omega$$
Show that the Fourier {\it series} of $F(\omega) = {\cal F}[f](\omega)$ on 
$[-L,L]$ is:
$$F(\omega) \sim {1\over 2L} \sum_{n=-\infty}^\infty
f\left({n\pi\over L}\right) e^{i n\pi\omega \over L}$$
(see section 3.6 in Haberman, or the class notes, 
for the notation of the complex form of the Fourier series).
\item{B)}  Let $H_L(\omega)$ be the periodic function defined by 
the Fourier series in part A.  Explain why
\medskip
\centerline{$F(\omega)$ ``$=$'' $H_L(\omega) \Pi_L(\omega),$}
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\item{}
so
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\centerline{$F(\omega)$ ``$=$'' ${1\over 2L}\sum_{n=-\infty}^\infty 
f\left({n\pi\over L}\right) e^{i n\pi\omega \over L} \Pi_L(\omega)$}
\medskip
\item{}
where $\Pi_L(\omega) = 1$ if $|\omega| \le L$ and $= 0$ otherwise.
Why are several equalities here written with quotation marks?  (These
are several technical details that are being overlooked for simplicity!)
\item{C)} Derive equation (4) by taking the Inverse Fourier Transform
of both sides of the last equation.  
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\centerline{\it Have a wonderful summer!}
\bye
